1.
a) From Figure 1.14, a value of a = .58 gives a value of u very close to 1.0.
Solving x = u/r = 1/.12 = 8.33 years. The optimal size of each addition is
xD = (8.33)(5,000) = 41,650 tons.
b) f(y) = .0205y.58 = (.0205)(41,650).58 = 9.79.
Hence, the cost of each new addition is $9.79 million.
c) At 12% interest the effective discount rate is e^(-0.12*8.33) = 0.368.
Hence, the present cost of the next four replacements is
2
3
4
9.79[.368 + (.368) + (.368) + (.368) ] = $5.6 million
2.
a) So = 3,500. Since the second week is the current position in time, it will be
relabeled t=0. Hence, when estimating So (interpreted as the intercept) we find
that 3,500 is the intercept because it’s the value of the ordinate when the abscissa
is zero.
Go = 1,500. Go is interpreted as the slope (rise/run = (3,500-2,000) / (2-1) =
1,500
Note that there might be other reasonable estimates for So. For example, if you
assume that 3,500 occurs in the middle of period 2, then it may be reasonable to
estimate So as 3,500 + Go/2 = 4,500. If you answered So = 2000 thinking that
you could use a τ that gives you a mathematically equivalent solution this will be
mathematically correct for the first estimate but the concept of So is incorrect.
Also, you will have to manipulate the formulas given to update the slope and
intercept.
b) Update S and G to consider the new data from period 3 (D3 = 4,000). Note that
period 2 has been relabeled as t = 0 in part a therefore period 3 is represented by
the index 1.
S1  D1  (1   )( S 0  G0 )  .2(4000)  (1  .2)(3500  1500)  4800
G1   (S1  S 0 )  (1   )G0  .2(4800  3500)  (1  .2)1500  1460
Ft ,t   S t  Gt
F1, 2  4800  1 *1460  6260
F1,3  4800  2 *1460  7720
3.
a)
b)
c)
Note that the inventory holding cost per hour of production in inventory is $100 * 5%. So
90 hours of ending inventory would cost 90*100*5% = $450.
The total overall cost is $62,500.
4.
= 40 (pounds/day)*30(days/month) = 1200 pounds/month
NOTE:
- I assumed 30 working days in a month.
- The holding cost is 2% which in this particular problem is the cost of opportunity.
Therefore, h = 2%*c. You hold 2% of the cost of each pound.
Part a)
Mark’s offer computations:
Step 1) Compute EOQ for each quantity discount price, and check feasibility:
Note: The formula for EOQ was recomputed to account for h=ic.
2 *150 *1200
 3721.04 lbs > 500 lbs (NOT FEASIBLE)
0.02 *1.3
2 *150 *1200

 3872.98 lbs > 1000 lbs (NOT FEASIBLE)
0.02 *1.2
2 *150 *1200

 4045.20 > 3000 lbs (FEASIBLE)
0.02 *1.1
Q (1) 
Q ( 2)
Q ( 3)
The largest discount is feasible -> 4045 is optimal.
Step 3) Compute the cost per unit time:
K
Q (i )*
( i )*
Gi Q
 (i )*  ci   ic i
2
Q


G3 4045.2  
150 * 1200
1.1 * 4045.2
 1.1 *1200  0.02
 1409
4045.2
2
If Gary is going to order from Mark he should order 4045 lbs.
Anne’s offer computations:
Step 1) Obtain average price for each order level:
C1ave = 1.25
C1ave = [1.25*700 + 1.05*(Q-700)] / Q
Step 1) Determine Gi(Q):
G1 Q  
150 *1200
1.25Q
 1.25 *1200  0.02
Q
2
1.25 * 700  1.05 * (Q  700)
*Q
150 *1200 1.25 * 700  1.05 * (Q  700)
Q
G2 Q  
[
] *1200  0.02
Q
Q
2
Step 2) Find Q* for each order level:
For incremental level 1 we use the EOQ formula for h=ic derived earlier.
Q (1) 
2 *150 *1200
 3794.73 lbs > 700 lbs (NOT FEASIBLE)
0.02 *1.25
For the second incremental level we must derive the EOQ formula:
1.25 * 700  1.05 * (Q  700)
*Q
150 *1200 1.25 * 700  1.05 * (Q  700)
Q
G2 Q  
[
] *1200  0.02
Q
Q
2
150  875  1.05Q  735)
[
] *1200  0.01 * (875  1.05Q  735)
Q
290 *1200

 1.05 *1200  1.4  .0105Q
Q
G2 (Q)
 1 * (290 *1200)
0
 .0105
Q
Q2
(290 *1200)
 .0105
Q2
Q ( 2)* 
290 *1200
, alternatively
.0105
Q ( 2)* 
2 *1200 * (150  700 *1.25  700 *1.05)
 5756.98  5757 (FEASIBLE)
.02 *1.05
Step 3) Compute cost of each quantity for feasible quantities:
150 *1200 1.25 * 700  1.05 * (5757  700)
G2 5757  
[
] *1200  0.02
5757
5757
1.25 * 700  1.05 * (5757  700)
* 5757
5757
2
G2 5757  31.26  1289.18  61.85  1382.30
Note that this cost is lower than Mark’s cost. So Gary should order from Anne.
Part c)
As shown in part a and b, Gary will order large amounts at once, and thus order very
infrequently. Given this frequency of ordering, I would probably not eat a hamburger at
his restaurant. The importance of this question is to determine the frequency of the
orders, not to figure out who’s vegetarian in our class.